# Newton's Law of Temperature Change Differential

1. Oct 11, 2012

### Manni

I'm a bit confused as to what I do next with this problem.

Consider the initial value problem $\frac{dy}{dx}$ = a(y-b) where y(0) = y0

With a > 0, b = 0, it represents exponential growth, while a < 0, b = 0 gives exponential decay. With y = T, a = -k, k = TA and y0=T0 it gives Newton's Law of Temperature Change,

$\frac{dy}{dx}$ = -k(T - TA)

Solve the given model and give a qualitative sketch for each of the cases 1) a > 0 and 2) a < 0. You may assume b ≥ 0. Include y0= b as one typical solution for each.

So what I did was I applied the integral to both sides of $\frac{dy}{dx}$ = -k(T - TA) yielding,

1 = -k ∫ (T - TA)

But I don't know how to integrate TA

Last edited: Oct 11, 2012
2. Oct 11, 2012

### HallsofIvy

Staff Emeritus
You can't integrate both sides because you have no differential on the right. Part of your problem is that you say y= T but then leave y in the equation.
Write the equation as $\frac{dT}{dx}= -k(T- T_A)$, then as
$$\frac{dT}{T- T_A}= -kdx$$
and integrate. TA is a constant. To integrate on the left, let u= T- TA so that du= dT.

3. Oct 11, 2012

### Ray Vickson

I think you are mis-reading the problem. It says to solve the given model, and to look at the cases a > 0 and a < 0, etc. To me, that says solve dy/dx = a*(y-b) [= the given model], so there is no "T" involved.

RGV

4. Oct 11, 2012

### Manni

Oh, so we would integrate the constant b as a normal constant correct? I.e. b*x?

5. Oct 11, 2012

### HallsofIvy

Staff Emeritus
Not when it is in the denominator! Use the substitution u= T- TA as I suggested.

6. Oct 11, 2012

### Manni

Oh I understand, thank you!

7. Oct 11, 2012

### Manni

Wait, it's dT/dt not dT/dx?

8. Oct 11, 2012

### Manni

The only thing I'm not sure of is if it the question is asking for solutions. If it is can I put the dy/dx equation in the standard differential form,

dy/dx + P(x) = Q(x)

Then, I can determine the integrating factor and solve that way. I was initially thinking of doing that, would it work?

Last edited: Oct 11, 2012